Min : Do x ; y không âm , \(x^2+y^2=1\) \(\Rightarrow\left|x\right|;\left|y\right|\le1\)
\(\Rightarrow0\le x;y\le1\)
\(\Rightarrow xy\ge0;x\ge x^2;y\ge y^2\)
\(P=\sqrt{1+2x}+\sqrt{1+2y}\)
\(\Rightarrow P^2=1+2x+1+2y+2\sqrt{1+2x+2y+2xy}\)
\(\ge2+2.1+2\sqrt{1+2.1+0}=4+2\sqrt{3}\)
\(\Rightarrow P\ge\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)
Max : Áp dụng BĐT phụ : \(\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\) , ta có :
\(P=\sqrt{1+2x}+\sqrt{1+2y}\le\sqrt{2\left(1+2x+1+2y\right)}\)
\(=\sqrt{2\left[2+2\left(x+y\right)\right]}\le\sqrt{2\left[2+2\sqrt{2\left(x^2+y^2\right)}\right]}=\sqrt{2\left(2+2\sqrt{2}\right)}=\sqrt{4+4\sqrt{2}}=2\sqrt{1+\sqrt{2}}\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=\frac{1}{\sqrt{2}}\)